# ^3√80 in simplest Radical form?

Feb 22, 2018

$2 \sqrt{10}$

#### Explanation:

I'm guessing that the problem is a cube root? Let me know if it's not. Assuming it's a cube root:

You need to look for perfect cube factors of 80. Perfect cubes are 8, 27, 64, 125, etc. Since 8 is a factor of $\sqrt{80}$, we can re-write the cube root like this:

$\sqrt{80} = \sqrt{8} \times \sqrt{10}$

Now simplify the perfect cube:

$\sqrt{8} \times \sqrt{10} = 2 \times \sqrt{10} = 2 \sqrt{10}$

Feb 22, 2018

$3 \setminus \sqrt{80} = 12 \setminus \sqrt{5}$

Or

$\sqrt{80} = 2 \sqrt{10}$

#### Explanation:

$3 \setminus \sqrt{80}$

$= 3 \setminus \sqrt{2 \setminus \times 2 \setminus \times 2 \setminus \times 2 \setminus \times 5}$

Apply the exponent rule to get:

$= 3 \setminus \sqrt{{2}^{2} \setminus \times {2}^{2} \setminus \times 5}$

$= 3 \setminus \cdot \setminus \sqrt{{2}^{2}} \setminus \cdot \setminus \sqrt{{2}^{2}} \setminus \sqrt{5}$

Cancel out the radical to get:

$= 3 \setminus \cdot 2 \setminus \cdot 2 \setminus \sqrt{5}$

Simplify:

$= 12 \sqrt{5}$

If you input is $\sqrt{80}$ , then the answer would be:

$= 2 \sqrt{10}$